Question

A circuit consists of two very low-resistance rails separated by 0.300 m, with a 50.0 ohm resistor connected across them at one end and a conducting, moveable bar at the other end. The circuit is in a uniform 0.100 T magnetic field that is parallel to the area vector of the circuit. The bar is moving at a constant velocity such that the resistor dissipates 1.50 W of power. What is the speed of the bar

Answer 1

Answer:

v = 288.67 m/s

Explanation:

Given that:

length of the separation I = 0.3

Resistance R = 50.0 ohms

Magnetic field B = 0.100 T

Power dissipated = 1.50 W

In a simple circuit, The Emf  voltage of the power dissipated can be determined via the expression:

Power =$\frac{V^2}{R}$

1.5 = $\frac{V^2}{50}$

$V^2 = 1.5*50$

$V^2 = 75$

$V = \sqrt{75}$

V = 8.660 V

From the Electromotive Force Emf ; we can calculate the speed of the bar

i.e Emf = B×v×l

8.660 = 0.1 × v × 0.3

v = $\frac {8.66}{0.1*0.3}$

v = 288.67 m/s